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3 years ago

A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

Physics

1 answer:

ANTONII [103]3 years ago

6 0

Answer:

$11.3 \Omega$

Explanation:

We can find the resistance of the wire by using Ohm's law:

$V=RI$

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

$R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega$

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Read 2 more answers

a solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of

Natalija [7]

Complete question:

A solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of 6.75 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

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(a) the energy density of the magnetic field inside the solenoid is 50.53 J/m³

(b) the total energy stored in the magnetic field is 0.121 J

Explanation:

Given;

length of the solenoid, L = 98.6 cm = 0.986 m

cross-sectional area of the solenoid, A = 24.3 cm² = 24.3 x 10⁻⁴ m²

number of turns of the solenoid, N = 1310 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μ₀nI

B = μ₀(N/L)I

Where;

μ₀ is permeability of free space, = 4π x 10⁻⁷ m/A

$B = \frac{4\pi*10^{-7}*1310*6.75}{0.986} \\\\B = 0.01127 \ T$

(a) Calculate the energy density of the magnetic field inside the solenoid

$u = \frac{B^2}{2 \mu_o}\\\\u = \frac{(0.01127)^2}{2*4\pi *10^{-7}} \\\\u = 50.53 \ J/m^3$

(b) Find the total energy stored in the magnetic field

U = uV

U = u (AL)

U = 50.53 (24.3 x 10⁻⁴ x 0.986)

U = 0.121 J

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